let . This function has five roots on the interval . I have to use the bisection method on this interval and figure out what root is located.
I'm not entirely sure I know how to use this method correctly..
Here it goes,
First, since and , there exists a in between those two points such that ..
Now would you divide the interval in two?

March 3rd 2010, 08:48 PM

pickslides

This is quite a strange function to use with this method as it is fully factored and gives you the roots.

Anyhow you choose the interval and find the signs of and
Now bisect the interval to make 2 smaller intervals and find the sign of

We know so we choose the smaller intervals with opposite signs and , bisect again and so on until the interval is small enough.

March 3rd 2010, 09:05 PM

frenchguy87

Quote:

Originally Posted by pickslides

This is quite a strange function to use with this method as it is fully factored and gives you the roots.

Anyhow you choose the interval and find the signs of and
Now bisect the interval to make 2 smaller intervals and find the sign of

We know so we choose the smaller intervals with opposite signs and , bisect again and so on until the interval is small enough.

Ok that's kind of what I thought... and yea it gives you all the roots but they want you to find the root that will be found after using that method, which apparently ends up being

March 3rd 2010, 09:26 PM

tn11631

Quote:

Originally Posted by frenchguy87

let . This function has five roots on the interval . I have to use the bisection method on this interval and figure out what root is located.
I'm not entirely sure I know how to use this method correctly..
Here it goes,
First, since and , there exists a in between those two points such that ..
Now would you divide the interval in two?

Wow thats crazy i just took an exam today with almost that same exact problem just a little difference in like one of the numbers. Anyway the way I did it was that I just kept applying the bisection method where a=0 b=7 and you have to find thier signs and all that, but then when you plug in p_n it either matches the sign of a or b, however you keep doing it until f(p)=0. However in your case you can kind of see where its going and it keeps getting closer and closer to 1 so your root that will be determined would be 1. I'm not sure if its all the correct but I mean it worked for me on both my hw's and the practice exam. Hope this helped

March 3rd 2010, 10:18 PM

frenchguy87

Quote:

Originally Posted by tn11631

Wow thats crazy i just took an exam today with almost that same exact problem just a little difference in like one of the numbers. Anyway the way I did it was that I just kept applying the bisection method where a=0 b=7 and you have to find thier signs and all that, but then when you plug in p_n it either matches the sign of a or b, however you keep doing it until f(p)=0. However in your case you can kind of see where its going and it keeps getting closer and closer to 1 so your root that will be determined would be 1. I'm not sure if its all the correct but I mean it worked for me on both my hw's and the practice exam. Hope this helped

How would you formally write the end of the proof, whenever you have a small enough interval?
I am down to the interval, so how would I go one to say that the root must be 1?

March 3rd 2010, 10:36 PM

Drexel28

Quote:

Originally Posted by frenchguy87

How would you formally write the end of the proof, whenever you have a small enough interval?
I am down to the interval, so how would I go one to say that the root must be 1?

I don't understand. Are you to play dumb? Clearly and . So clearly the zero given must be one.

March 3rd 2010, 11:06 PM

frenchguy87

Quote:

Originally Posted by Drexel28

I don't understand. Are you to play dumb? Clearly and . So clearly the zero given must be one.